1
$\begingroup$

Or is the kinetic energy of the electrons not counted towards the thermal energy? (In other words only the energy of the lattice structure is counted)

$\endgroup$
  • $\begingroup$ connected physics.stackexchange.com/questions/80711/… . The potential and kinetic energy of electrons bound in the potential around the nucleus of the atoms composing the material is part of the internal energy of the system ( which ultimately includes the E=mc**2 of the masses) $\endgroup$ – anna v Mar 12 '14 at 6:52
  • $\begingroup$ @annav So the kinetic energy of the electrons is included? $\endgroup$ – dfg Mar 12 '14 at 16:20
  • $\begingroup$ No. It is within the packaged atoms, part of the internal energy . The electrons are in quantized non changeable energy levels about the nucleus. The atoms as a whole have vibrational degrees of freedom in the lattice. As far as the definition of heat is concerned, which is a thermodynamic concept, but has been described by a statistical mechanics formulation, the atoms are non divisible, what their name says. en.wikipedia.org/wiki/Statistical_mechanics $\endgroup$ – anna v Mar 12 '14 at 17:03
1
$\begingroup$

If I understand correctly, the question is: is the bulk motion of electrons taken into account in the thermal energy?

Thermal energy is energy that is distributed among the degrees of freedom of your system. "If a system contains N molecules, each with f degrees of freedom, and there are no other (non quadratic) temperature-dependent forms of energy, then its total thermal energy is": $$U_{thermal}=Nf\frac{1}{2}kT$$ This is temperature-dependent energy. Now lets look at the degrees of freedom among which it could be distributed (depending which modes are available).. it includes $\frac{1}{2}mv_x^2$, $\frac{1}{2}mv_y^2$,...,$\frac{1}{2}\omega_x^2$ etc.

Now the equipartition theorem states: "At temperature T, the average energy of any quadratic degree of freedom is" $$\frac{1}{2}kT$$ This kinetic energy due to the temperature of a substance is distributed equally among the available degrees of freedom. I think the kinetic energy you are referring to is bulk motion, which has nothing to do with the equilibrium statistics of a system.

The subtle aspect here is that a system can be in thermodynamic equilibrium with its constituents obeying bolztmann statistics while the system as a whole may be moving. Try this for a thought experiment; imagine moving the "microcanonical ensemble" at a constant velocity. Are the bolztmann statistics affected if you are in the reference frame of the moving "box of particles"?

Ref: Schroeder page 15 <3.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.